Solve (2x)/3 + (x ^ 2 - 3)/4 = (x ^ 2)/6 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$\frac{2x}{3}+\frac{x^{2}-3}{4}=\frac{x^{2}}{6}$$

Solve for x

$x=-9$
$x=1$

Steps Using Factoring

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring

Multiply both sides of the equation by $12$, the least common multiple of $3,4,6$.

$$4\times 2x+3\left(x^{2}-3\right)=2x^{2}$$

Multiply $4$ and $2$ to get $8$.

$$8x+3\left(x^{2}-3\right)=2x^{2}$$

Use the distributive property to multiply $3$ by $x^{2}-3$.

$$8x+3x^{2}-9=2x^{2}$$

Subtract $2x^{2}$ from both sides.

$$8x+3x^{2}-9-2x^{2}=0$$

Combine $3x^{2}$ and $-2x^{2}$ to get $x^{2}$.

$$8x+x^{2}-9=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}+8x-9=0$$

To solve the equation, factor $x^{2}+8x-9$ using formula $x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right)$. To find $a$ and $b$, set up a system to be solved.

$$a+b=8$$ $$ab=-9$$

Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product $-9$.

$$-1,9$$ $$-3,3$$

Calculate the sum for each pair.

$$-1+9=8$$ $$-3+3=0$$

The solution is the pair that gives sum $8$.

$$a=-1$$ $$b=9$$

Rewrite factored expression $\left(x+a\right)\left(x+b\right)$ using the obtained values.

$$\left(x-1\right)\left(x+9\right)$$

To find equation solutions, solve $x-1=0$ and $x+9=0$.

$$x=1$$ $$x=-9$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by $12$, the least common multiple of $3,4,6$.

$$4\times 2x+3\left(x^{2}-3\right)=2x^{2}$$

Multiply $4$ and $2$ to get $8$.

$$8x+3\left(x^{2}-3\right)=2x^{2}$$

Use the distributive property to multiply $3$ by $x^{2}-3$.

$$8x+3x^{2}-9=2x^{2}$$

Subtract $2x^{2}$ from both sides.

$$8x+3x^{2}-9-2x^{2}=0$$

Combine $3x^{2}$ and $-2x^{2}$ to get $x^{2}$.

$$8x+x^{2}-9=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}+8x-9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $x^{2}+ax+bx-9$. To find $a$ and $b$, set up a system to be solved.

$$a+b=8$$ $$ab=1\left(-9\right)=-9$$

Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product $-9$.

$$-1,9$$ $$-3,3$$

Calculate the sum for each pair.

$$-1+9=8$$ $$-3+3=0$$

The solution is the pair that gives sum $8$.

$$a=-1$$ $$b=9$$

Rewrite $x^{2}+8x-9$ as $\left(x^{2}-x\right)+\left(9x-9\right)$.

$$\left(x^{2}-x\right)+\left(9x-9\right)$$

Factor out $x$ in the first and $9$ in the second group.

$$x\left(x-1\right)+9\left(x-1\right)$$

Factor out common term $x-1$ by using distributive property.

$$\left(x-1\right)\left(x+9\right)$$

To find equation solutions, solve $x-1=0$ and $x+9=0$.

$$x=1$$ $$x=-9$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by $12$, the least common multiple of $3,4,6$.

$$4\times 2x+3\left(x^{2}-3\right)=2x^{2}$$

Multiply $4$ and $2$ to get $8$.

$$8x+3\left(x^{2}-3\right)=2x^{2}$$

Use the distributive property to multiply $3$ by $x^{2}-3$.

$$8x+3x^{2}-9=2x^{2}$$

Subtract $2x^{2}$ from both sides.

$$8x+3x^{2}-9-2x^{2}=0$$

Combine $3x^{2}$ and $-2x^{2}$ to get $x^{2}$.

$$8x+x^{2}-9=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.

$$x^{2}+8x-9=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $8$ for $b$, and $-9$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-9\right)}}{2}$$

Square $8$.

$$x=\frac{-8±\sqrt{64-4\left(-9\right)}}{2}$$

Multiply $-4$ times $-9$.

$$x=\frac{-8±\sqrt{64+36}}{2}$$

Add $64$ to $36$.

$$x=\frac{-8±\sqrt{100}}{2}$$

Take the square root of $100$.

$$x=\frac{-8±10}{2}$$

Now solve the equation $x=\frac{-8±10}{2}$ when $±$ is plus. Add $-8$ to $10$.

$$x=\frac{2}{2}$$

Divide $2$ by $2$.

$$x=1$$

Now solve the equation $x=\frac{-8±10}{2}$ when $±$ is minus. Subtract $10$ from $-8$.

$$x=-\frac{18}{2}$$

Divide $-18$ by $2$.

$$x=-9$$

The equation is now solved.

$$x=1$$ $$x=-9$$

Steps for Completing the Square

Multiply both sides of the equation by $12$, the least common multiple of $3,4,6$.

$$4\times 2x+3\left(x^{2}-3\right)=2x^{2}$$

Multiply $4$ and $2$ to get $8$.

$$8x+3\left(x^{2}-3\right)=2x^{2}$$

Use the distributive property to multiply $3$ by $x^{2}-3$.

$$8x+3x^{2}-9=2x^{2}$$

Subtract $2x^{2}$ from both sides.

$$8x+3x^{2}-9-2x^{2}=0$$

Combine $3x^{2}$ and $-2x^{2}$ to get $x^{2}$.

$$8x+x^{2}-9=0$$

Add $9$ to both sides. Anything plus zero gives itself.

$$8x+x^{2}=9$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.

$$x^{2}+8x=9$$

Divide $8$, the coefficient of the $x$ term, by $2$ to get $4$. Then add the square of $4$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+8x+4^{2}=9+4^{2}$$

Square $4$.

$$x^{2}+8x+16=9+16$$

Add $9$ to $16$.

$$x^{2}+8x+16=25$$

Factor $x^{2}+8x+16$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+4\right)^{2}=25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+4\right)^{2}}=\sqrt{25}$$

Simplify.

$$x+4=5$$ $$x+4=-5$$

Subtract $4$ from both sides of the equation.

$$x=1$$ $$x=-9$$